solve this system using a rref matrix: 3x − 2z = −3
−2x + z = −2
− z = 2

We are given the following system of equations:

\[
\begin{aligned}
1. \ 3x - 2z &= -3 \\
2. \ -2x + z &= -2 \\
3. \ -z &= 2
\end{aligned}
\]

### Step 1: Rewrite the system in augmented matrix form.

We can rewrite the system as:

\[
\begin{pmatrix}
3 & 0 & -2 & | & -3 \\
-2 & 0 & 1 & | & -2 \\
0 & 0 & -1 & | & 2
\end{pmatrix}
\]

### Step 2: Perform row operations to obtain the Reduced Row Echelon Form (RREF).

We will transform this matrix step-by-step into the RREF.

#### Row 3 operation:
First, multiply row 3 by \(-1\) to make the pivot element in the third row a positive 1.

\[
R_3 \to -1 \times R_3
\]

\[
\begin{pmatrix}
3 & 0 & -2 & | & -3 \\
-2 & 0 & 1 & | & -2 \\
0 & 0 & 1 & | & -2
\end{pmatrix}
\]

#### Row 2 operation:
Now eliminate the third element in row 2 by adding row 3 multiplied by \(-1\) to row 2.

\[
R_2 \to R_2 + (-1) \times R_3
\]

\[
\begin{pmatrix}
3 & 0 & -2 & | & -3 \\
-2 & 0 & 0 & | & 0 \\
0 & 0 & 1 & | & -2
\end{pmatrix}
\]

#### Row 1 operation:
Now eliminate the third element in row 1 by adding row 3 multiplied by \(2\) to row 1.

\[
R_1 \to R_1 + 2 \times R_3
\]

\[
\begin{pmatrix}
3 & 0 & 0 & | & -7 \\
-2 & 0 & 0 & | & 0 \\
0 & 0 & 1 & | & -2
\end{pmatrix}
\]

#### Row 1 operation (dividing by 3):
To get the leading coefficient in row 1 to be 1, divide row 1 by 3.

\[
R_1 \to \frac{1}{3} \times R_1
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -\frac{7}{3} \\
-2 & 0 & 0 & | & 0 \\
0 & 0 & 1 & | & -2
\end{pmatrix}
\]

#### Row 2 operation (dividing by -2):
To get the leading coefficient in row 2 to be 1, divide row 2 by \(-2\).

\[
R_2 \to \frac{1}{-2} \times R_2
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -\frac{7}{3} \\
1 & 0 & 0 & | & 0 \\
0 & 0 & 1 & | & -2
\end{pmatrix}
\]

### Step 3: Interpretation of the Solution

From the final matrix, we can now read off the solution:

\[
x = -\frac{7}{3}, \quad z = -2
\]

The variable \(y\) does not appear in the equations, meaning it is a free variable, so the system allows \(y\) to take any real value. Therefore, the solution can be written as:

\[
x = -\frac{7}{3}, \quad z = -2, \quad y \in \mathbb{R}
\]

### Would you like more details or have any questions?

Here are 5 related questions:
1. How do you interpret free variables in a solution set?
2. What are the key advantages of using RREF for solving systems?
3. Can every system of linear equations be solved using RREF?
4. How does the number of free variables relate to the solution space?
5. What are the computational steps to obtain the inverse of a matrix using row operations?

**Tip:** When solving systems of equations, RREF is particularly useful for identifying free variables and ensuring a unique or general solution.

solve this system using a rref matrix: 3x − 2z = −3, −2x + z = −2, −z = 2

This tutorial explains how to solve a system of linear equations using the Reduced Row Echelon Form (RREF) method. The system includes equations like 3x - 2z = -3, -2x + z = -2, and -z = 2. Suitable for students in Grades 10-12, this problem covers matrix operations and systems of equations.

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